1991 AMC 12 Problem 16

Attempt Problem 16 of the 1991 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1991 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

16.

One hundred students at Century High School participated in the AHSME last year, and their mean score was 100.100. The number of non-seniors taking the AHSME was 50%50\% more than the number of seniors, and the mean score of the seniors was 50%50\% higher than that of the non-seniors. What was the mean score of the seniors?
100100
112.5112.5
120120
125125
150150
Answer: D
Concepts:weighted averagepercentlinear equation
Small hint:
First determine the numbers of seniors and non-seniors
Big hint:
Let the non-senior mean be xx and form a weighted-average equation
Solution:
If there are ss seniors, then there are 1.5s1.5s non-seniors, so 2.5s=1002.5s=100 and s=40.s=40. Let the non-senior mean be x;x; the senior mean is 1.5x.1.5x. The total score equation is 60x+40(1.5x)=100(100), 60x+40(1.5x)=100(100), so 120x=10000,120x=10000, and the senior mean is 1.5x=125.1.5x=125.
Thus the correct answer is D.
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